A379402 Rectangular array, read by descending antidiagonals: the Type 2 runlength index array of A039701 (primes mod 3); see Comments.

1, 2, 9, 3, 11, 15, 4, 16, 18, 54, 5, 21, 23, 58, 91, 6, 32, 36, 102, 110, 205, 7, 37, 39, 129, 160, 272, 194, 8, 40, 46, 161, 167, 419, 271, 139, 10, 47, 55, 174, 238, 499, 416, 260, 86, 12, 56, 73, 245, 273, 597, 496, 359, 257, 357, 13, 67, 96, 274, 292

Offset: 1

Author

Clark Kimberling, Jan 15 2025

Keywords

• nonn
• tabl

Comments

We begin with a definition of Type 2 runlength array, V(s), of any sequence s for which all the runs referred to have finite length:

Suppose s is a sequence (finite or infinite), and define rows of V(s) as follows:

(row 0) = s

(row 1) = sequence of last terms of runs in (row 0); c(1) = complement of (row 1) in (row 0)

For n>=2,

(row n) = sequence of last terms of runs in c(n-1); c(n) = complement of (row n) in (row n-1),

where the process stops if and when c(n) is empty for some n.

***

The corresponding Type 2 runlength index array, The runlength index array, VI(s) is now contructed from V(s) in two steps:

(1) Let V*(s) be the array obtaining by repeating the construction of V(s) using (n,s(n)) in place of s(n).

(2) Then VI(s) results by retaining only n in V*.

Thus, loosely speaking, (row n) of VI(s) shows the indices in s of the numbers in (row n) of V(s).

The array VI(s) includes every positive integer exactly once.

***

Regarding the present array, each row of U(s) splits a sequence of primes according to remainder modulo 3; e.g., in row 2, the remainders of primes in positions 9,11,16,21,32,37,40,47 are 2,1,2,1,2,1,2,1, respectively.

Conjecture: every column is eventually increasing.

Examples

			Corner:
      1    2    3    4      5     6     7     8    10    12    13    14
      9   11   16   21     32    37    40    47    56    67    71    74
     15   18   23   36     39    46    55    73    96    99   107   111
     54   58  102  129    161   174   245   274   311   326   423   515
     91  110  160  167    238   273   292   321   420   508   598   621
    205  272  419  499    597   618   703   733   813   835   896   932
    194  271  416  496    576   617   702   730   776   834   989  1128
    139  260  359  489    699   713   771   831   988  1127  1173  1190
     86  257  358  464    698   830   987  1124  1164  1185  1251  1298
    357  461  697  829    942  1107  1412  1498  1717  2059  2138  2179
    356  438  889  1062  1714  2046  2137  2176  2551  2820  2927  3270
    291  437  882  1055  1711  2033  2550  2741  2926  3269  3699  3918
Starting with s = A039701, we have for U*(s):
(row 1) = ((1,1), (2,0), (3,2), (4,2), (5,2), (6,1), (7,2), (8,1), (10,2), ...)
c(1) = ((9,2), (11,1), (15,2), (16,2), (18,1), (21,1), (23,1), (32,2), ...)
(row 2) = ((9,2), (11,1), (16,2), (21,1), (36,1), ...)
c(2) = ((15,2), (37,1), ...)
(row 3) = ((15,2), (18,1), (23,2), ...)
so that UI(s) has
(row 1) = (1,2,3,4,5,6,7,8,10,12,13, ...)
(row 2) = (9,11,16.21,32, ...)
(row 3) = (15,18,23,...)
		

Crossrefs

Cf. A000040, A039701, A379046, A379401, A379403, A379404.

Programs

Mathematica

r[seq_] := seq[[Flatten[Position[Append[Differences[seq[[All, 1]]], 1], _?(# != 0 &)]], 2]];  (* Type 2 *)
row[0] = Mod[Prime[Range[4000]], 3];(* A039701 *)
row[0] = Transpose[{#, Range[Length[#]]}] &[row[0]];
k = 0; Quiet[While[Head[row[k]] === List, row[k + 1] = row[0][[r[
     SortBy[Apply[Complement, Map[row[#] &, Range[0, k]]], #[[2]] &]]]]; k++]];
m = Map[Map[#[[2]] &, row[#]] &, Range[k - 1]];
p[n_] := Take[m[[n]], 12]
t = Table[p[n], {n, 1, 12}]
Grid[t]  (* array *)
w[n_, k_] := t[[n]][[k]];
Table[w[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten  (* sequence *)
(* Peter J. C. Moses, Dec 04 2024 *)